# Prove: Every compact metric space is separable

How to prove that Every compact metric space is separable$?$

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Hint: Consider the countable family of coverings $\mathcal U_n=\left\{B\left(x,\frac1n\right)\mid x\in X\right\}$ for all $n$, use compactness to distill a countable family of points, and show it is dense.

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Thanks for the hint, i got it. But, i feel it difficult to find this kind of open covering. There are a lot of other open coverings ; is there any trick for how to know which one would help? –  Aang Sep 4 '12 at 6:16
@Avatar: All of them. –  Asaf Karagila Sep 4 '12 at 6:20
The easiest open set in a metric space are balls; the easiest way to ensure that all points are covered is to use all points as centers; the easiest way to ensure best "granularity" is to allow arbitrarily small radii. –  Hagen von Eitzen Sep 4 '12 at 6:22
@Asaf Karagila: You exhibit a countable set of coverings, whic might be confusing. The straightforward covering $\{B(x,r)|x \in X, r>0\}$ should suffice. –  Hagen von Eitzen Sep 4 '12 at 6:24
@Avatar: But why would you care about other open covers? All you want is a countable dense set, and Asaf’s argument gives it to you. –  Brian M. Scott Sep 4 '12 at 6:25

Recall that every infinite subset $E$ of a compact metric space $X$ has a limit point in $X$. Now define a recursive process as follows: Fix some $\delta > 0$ and pick an $x_1 \in X$. Now having defined $x_1,\ldots,x_n$, if possible choose $x_{n+1}$ such that $d(x_{n+1},x_i) \geq \delta$ for all $1 \leq i \leq n$. Now we claim that this process must terminate. For suppose it does not. Consider $$E = \{x_1,x_2,x_3,\ldots \}.$$

Now by assumption of $X$ being compact, $E$ has a limit point $a \in X$ say. But then $B_\frac{\delta}{2}(a)$ can only contain at most one point of $E$, a contradiction. It follows that $X$ can be covered by finitely many neighbourhoods of radius $\delta$ about some points $x_1,\ldots,x_n$. Now what happens if you consider all balls of radius $\frac{1}{m}$ about each $x_i$ for $1 \leq i \leq n$? This will be countable, but why will it be dense in $X$? $m$ by the way runs through all positive integers.

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Just a minor quibble: $B_\delta (a)$ could contain more than one point of $E$. $B_{\delta / 2} (a)$ on the other hand... –  Arthur Fischer Sep 4 '12 at 11:16
@ArthurFischer I always get mixed up with the radius and diameter :D –  user38268 Sep 4 '12 at 13:07
The $n$ depends on the $m$. We get different points for each radius $1/m$. –  Stefan Hamcke Feb 9 at 22:59
@ArthurFischer Why does "every infinite sub-set $E$ of a compact metric space $X$ has a limit point in $X$"? –  Eric_ Oct 19 at 18:30